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Given $X_1$, $X_2$ are independent EXP(1) random variables, how do I compute the following expectations:

  • $E[X_1 | X_1 + X_2]$

  • $P(X_1 > 3 | X_1 + X_2)$

  • $E[X_1 | min(X_1, t)]$

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  • $\begingroup$ Is this a homework question? If so, it should be flagged as such. $\endgroup$ – Placidia Dec 9 '12 at 1:35
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I'll do the first one. Please, try to find similar solutions to the others. We know that $X_1$ and $X_2$ are IID (their specific distributions don't matter yet). Hence, by symmetry we have $$ \textrm{E}[X_1\mid X_1+X_2] = \textrm{E}[X_2\mid X_1+X_2] \quad \textrm{a.s.} $$ Therefore, using the properties of the conditional expectation, we find $$ \textrm{E}[X_1\mid X_1 + X_2] = \frac{1}{2} \Bigg( \textrm{E}[X_1\mid X_1+X_2] + \textrm{E}[X_2\mid X_1+X_2] \Bigg) $$ $$ = \frac{1}{2} \textrm{E}[X_1 + X_2 \mid X_1+X_2] =\frac{X_1+X_2}{2} \quad \textrm{a.s.} $$ Since $X_1$ and $X_2$ are independent $\textrm{Exp}(1)\sim\textrm{Gamma}(1,1)$, we know that $$ X_1+X_2\sim\textrm{Gamma}(2,1) \, , $$ yielding $$ \textrm{E}[X_1\mid X_1+X_2] \sim \textrm{Gamma}(2,1/2) \, . $$ It's always good to make a quick verification. By the tower property, we know that it must be the case that $$ \textrm{E}[\textrm{E}[X_1\mid X_1+X_2]]=\textrm{E}[X_1] \, , $$ but from what we've just found, we have $\textrm{E}[\textrm{E}[X_1\mid X_1+X_2]]=1$, and also $\textrm{E}[X_1]=1$.

So the expectations match as expected, which is not bad.

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    $\begingroup$ Very pretty! I appreciate the separation into an idea that is distribution-independent, followed by specialization to a particular distribution. Although the other two parts of the question might not yield to the same strategy, your example nevertheless should inspire us to look for similarly clever ways to simplify them, too. $\endgroup$ – whuber Dec 10 '12 at 20:28
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    $\begingroup$ Rush (Zen), you can merge multiple accounts using the form at stats.stackexchange.com/help/user-merge. (Mods can no longer do this themselves because we have no mechanism to check identities.) $\endgroup$ – whuber Mar 22 '13 at 19:48
  • $\begingroup$ May I know to calculate $E[X_1 | min(X_1, t)]$, do we have to go through the standard procedure, that is, find the joint distribution of $(X_1 , min(X_1, t))$ and do integration? $\endgroup$ – John Feb 15 '16 at 15:13

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